# Inverse Trigonometric Ratios for Arbitrary Values

## Introduction:

Suppose that \(\sin \theta = \frac{1}{2}\). It is easy to invert this and write,

\[\theta = {\sin ^{ - 1}}\left( {\frac{1}{2}} \right) = \frac{\pi }{6}\]

Similarly, \(\cos \theta = \frac{1}{2}\) can be inverted to write,

\[\theta = {\cos ^{ - 1}}\left( {\frac{1}{2}} \right) = \frac{\pi }{3}\]

Now, suppose that we have the equation \(\sin \theta = \frac{3}{5}\). How do we invert this equation and write the value of \(\theta \)? Since there is no **readily available** value of \(\theta \) (in our minds) for which \(\sin \theta = \frac{3}{5}\), we do the inversion by simply writing,

\[\theta = {\sin ^{ - 1}}\left( {\frac{3}{5}} \right)\]

We do not then write a specific value for \(\theta \) (unless asked for explicitly – in which case you will require trigonometric tables to locate the angle for which \(\sin \theta = \frac{3}{5}\)).

**✍Note:** The **actual** value of \({\sin ^{ - 1}}\left( {\frac{3}{5}} \right)\) is about 36.9^{0}. You can check this using calculator:

\[\sin \left( {{{36.9}^0}} \right) \approx 0.60\]

As another example, consider the equation \(\cos \theta = \frac{1}{4}\). Inverting this, we can write:

\[\theta = {\cos ^{ - 1}}\left( {\frac{1}{4}} \right)\]

**✍Note: **We can’t do much more than this without a calculator or trigonometric tables, since we can’t mentally calculate the angle whose cosine will be \(\frac{1}{4}\). However, a calculator will tell you that,

\[{\cos ^{ - 1}}\left( {\frac{1}{4}} \right) \approx {75.5^0}\]

## Solved Examples:

**Example 1:** Find all values of \(x\) such that \(\sin x = \frac{1}{3}\).

**Solution:** In the principal range of the \({\sin ^{ - 1}}\) operation, the correct value of \(x\) will be

\[x = {\sin ^{ - 1}}\left( {\frac{1}{3}} \right)\]

Observe the following figure, which shows the intersection of \(y = \frac{1}{3}\) with \(y = \sin x\):

The highlighted point corresponds to the value of \(x\) in the principal range, which we can write as \({\sin ^{ - 1}}\left( {\frac{1}{3}} \right)\) . Another value of \(x\) which we can observe from this figure is

\[x = \pi - {\sin ^{ - 1}}\left( {\frac{1}{3}} \right)\]

In the interval \(\left[ {0,2\pi } \right]\), these are the only two values of \(x\) for which \(\sin x = \frac{1}{3}\). Now, we can generalize and write other values of \(x\) for which \(\sin x = \frac{1}{3}\) (\(n\) lies in the set of integers):

\[\boxed{x = 2n\pi + {\sin ^{ - 1}}\left( {\frac{1}{3}} \right),2n\pi + \pi - {\sin ^{ - 1}}\left( {\frac{1}{3}} \right)}\]

**Example 2:** Find all values of \(x\) such that \(\cos x = - \frac{2}{5}\).

**Solution:**The principal range of the \({\cos ^{ - 1}}\) operation is the interval \(\left[ {0,\pi } \right]\). The following figure shows the intersection of \(y = - \frac{2}{5}\) with \(y = \cos x\):

In \(\left[ {0,\pi } \right]\), we see one intersection point. We can write the \(x\)-coordinate of this intersection point as

\[{x_p} = {\cos ^{ - 1}}\left( { - \frac{2}{5}} \right)\]

A calculator will tell you that this value is roughly around 113.6^{0}.

We can also see another possible solution in the interval \(\left[ {0,2\pi } \right]\). Recall that \(\cos x = \cos \left( {2\pi - x} \right)\). Thus, this second solution is

\[2\pi - {x_p} = 2\pi - {\cos ^{ - 1}}\left( { - \frac{2}{5}} \right)\]

Now, we can write the general set of solutions as follows (\(n\) lies in the set of integers):

\[\boxed{x = 2n\pi + {\cos ^{ - 1}}\left( { - \frac{2}{5}} \right),2n\pi + 2\pi - {\cos ^{ - 1}}\left( { - \frac{2}{5}} \right)}\]

**Challenge:** Find all values of \(x\) such that \(\sin x = - \frac{1}{4}\)

**⚡****Tip:** Use a similar approach as in example-1.